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SAT II Math II : Real and Complex Numbers

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Example Questions

Example Question #11 : Sat Subject Test In Math Ii

Evaluate:

$\left ( 7+3i\right )^{2}$

Possible Answers:

58 + 42i

40 - 42i

40 + 42i

Correct answer:

40 + 42i

Explanation:

Use the square of a sum pattern

$\left ( A+B\right ) = A^{2}+ 2AB + B^{2}$

where A = 7, B = 3i :

$\left ( 7+3i\right )^{2} = 7^{2}+ 2 \cdot 7 \cdot 3i + (3i)^{2}$

$= 7^{2}+ 2 \cdot 7 \cdot 3i + 3^{2}i^{2}$

= 49 + 42i + 9 (-1)

= 49 + 42i -9

= 40 + 42i

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Example Question #11 : Real And Complex Numbers

is a complex number; $\overline{x}$ denotes the complex conjugate of .

$x \cdot \overline{x} = 84$

Which of the following could be the value of ?

Possible Answers:

$9 - i \sqrt{3}$

Any of the numbers in the other four choices could be equal to .

$6 - 4 i \sqrt{3}$

$7 + i \sqrt{35}$

$8 + 2 i \sqrt{5}$

Correct answer:

Any of the numbers in the other four choices could be equal to .

Explanation:

The product of a complex number x = a+ bi and its complex conjugate $\overline{x} = a- bi$ is

$x \cdot \overline{x} = (a+ bi)(a-bi) = a^{2} + b^{2}$

Setting and accordingly for each of the four choices, we want to find the choice for which $x \cdot \overline{x} = a^{2} + b^{2} = 84$ :

$6 - 4 i \sqrt{3}$

$a = 6, b = - 4 \sqrt{3}$

$x \cdot \overline{x} = a^{2} + b^{2} = 6^{2}+ (-4 \sqrt{3})^{2} = 36 + 4 ^{2} (\sqrt{3} )^{2} = 36 + 16 \cdot 3 = 36 + 48 = 84$

$7 + i \sqrt{35}$

$a = 7, b = \sqrt{35}$

$x \cdot \overline{x} = a^{2} + b^{2} = 7^{2} + (\sqrt{35})^{2} = 49 + 35 = 84$

$8 + 2 i \sqrt{5}$

$a = 8 , b = 2 \sqrt{5}$

$x \cdot \overline{x} = a^{2} + b^{2} = 8^{2} + (2 \sqrt{5})^{2} = 64 + 2^{2} \cdot (\sqrt{5})^{2} = 64 + 4 \cdot 5 = 64 + 20 = 84$

$9 - i \sqrt{3}$

$a = 9 , b = - \sqrt{3}$

$x \cdot \overline{x} = a^{2} + b^{2} = 9^{2} + (- \sqrt{3})^{2} = 81 + 3 = 84$

For each given value of , $x \cdot \overline{x} = 84$ .

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Example Question #21 : Number Theory

$\overline{x}$ denotes the complex conjugate of .

If $x = 5 + i \sqrt{2}$ , then evaluate $x^{4} \cdot \overline{x} ^{4}$ .

Possible Answers:

2,500

279,841

390,609

390,641

531, 441

Correct answer:

531, 441

Explanation:

Applying the Power of a Product Rule:

$x^{4} \cdot \overline{x} ^{4}=(x \overline{x})^{4}$

The complex conjugate of an imaginary number x = a+ bi is $\overline{x} = a - bi$ ; the product of the two is

$x \overline{x} = a^{2} + b^{2}$

$x = 5 + i \sqrt{2}$ , so, setting $a = 5 , b = \sqrt{2}$ in the above pattern:

$x \overline{x} = 5^{2} + (\sqrt{2})^{2} = 25 + 2= 27$

Consequently,

$x^{4} \cdot \overline{x} ^{4}=(x \overline{x})^{4} = 27 ^{4} = 531, 441$

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Example Question #11 : Real And Complex Numbers

Let and be complex numbers. $\overline{x}$ and $\overline{y}$ denote their complex conjugates.

$\overline{x} \cdot \overline{y } = 7 + 8 i \sqrt{2}$

$y = 8 - i\sqrt{3}$

Evaluate .

Possible Answers:

$8 \sqrt{2}+ 7i$

$-7 + 8 i \sqrt{2}$

$8 \sqrt{2} - 7i$

$-7 - 8 i \sqrt{2}$

$7 - 8 i \sqrt{2}$

Correct answer:

$7 - 8 i \sqrt{2}$

Explanation:

Knowing the actual values of and is not necessary to solve this problem. The product of the complex conjugates of two numbers is equal to the complex conjugate of the product of the numbers; that is,

$\overline{xy }= \overline{x} \cdot \overline{y }$

We are given that $\overline{xy }= \overline{x} \cdot \overline{y } = 7 + 8 i \sqrt{2}$ . is therefore the conjugate of $\overline{xy }$ , or $7 - 8 i \sqrt{2}$ .

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Example Question #11 : Real And Complex Numbers

$\overline{x}$ denotes the complex conjugate of .

If $x = \frac{1}{10} - \frac{1}{5} i$ , then evaluate $\overline{x}^{2} - x^{2}$ .

Possible Answers:

$\frac{2}{25} i$

$\frac{1}{25}$

$-\frac{4}{25 }$

$-\frac{2}{25}$

$\frac{4}{25 }i$

Correct answer:

$\frac{2}{25} i$

Explanation:

By the difference of squares pattern,

$\overline{x}^{2} - x^{2} = (\overline{x}+ x) ( \overline{x} - x)$

If $x = \frac{1}{10} - \frac{1}{5} i$ , then $\overline{x} = \frac{1}{10} + \frac{1}{5} i$ .

Consequently:

$\overline{x} + x =\left ( \frac{1}{10} + \frac{1}{5} i \right ) + \left ( \frac{1}{10} - \frac{1}{5} i \right )$

$= \frac{1}{10}+ \frac{1}{10} + \frac{1}{5} i- \frac{1}{5} i$

$= \frac{1}{5}$

$\overline{x} - x =\left ( \frac{1}{10} + \frac{1}{5} i \right ) - \left ( \frac{1}{10} - \frac{1}{5} i \right )$

$= \frac{1}{10}- \frac{1}{10} + \frac{1}{5} i+ \frac{1}{5} i$

$= \frac{2}{5} i$

Therefore,

$\overline{x}^{2} - x^{2} = (\overline{x}+ x) ( \overline{x} - x) = \frac{1}{5} \cdot \frac{2}{5} i = \frac{2}{25} i$

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Example Question #22 : Number Theory

The fraction $\frac{1+i}{1-i}$ is equivalent to which of the following?

Possible Answers:

Undefined

$\frac{1}{i}$

Correct answer:

Explanation:

Start by multiplying the fraction by $\frac{1+i}{1+i}$ .

$\frac{1+i}{1-i}(\frac{1+i}{1+i})=\frac{1+2i+i^2}{1-i^2}$

Since i^2=-1 , we can then simplify the fraction:

$\frac{1+2i+i^2}{1-i^2}=\frac{2i}{2}=i$

Thus, the fraction is equivalent to .

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Example Question #23 : Number Theory

The fraction $\frac{5+3i}{2-2i}$ is equivalent to which of the following?

Possible Answers:

$\frac{1}{2}+2i$

$\frac{3}{2}+3i$

1+2i

$4-\frac{1}{2}i$

Correct answer:

$\frac{1}{2}+2i$

Explanation:

Start by multiplying both the denominator and the numerator by the conjugate of 2-2i , which is 2+2i .

$\frac{5+3i}{2-2i}(\frac{2+2i}{2+2i})=\frac{10+10i+6i+6i^2}{4-4i^2}$

Next, recall i^2=-1 , and combine like terms.

$\frac{10+10i+6i+6i^2}{4-4i^2}=\frac{10+16i+6(-1)}{4-4(-1)}=\frac{4+16i}{8}$

Finally, simplify the fraction.

$\frac{4+16i}{8}=\frac{1}{2}+2i$

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Example Question #24 : Number Theory

Evaluate $i^{99}$

Possible Answers:

-99i

99i

Correct answer:

Explanation:

To evaluate a power of , divide the exponent by 4 and note the remainder.

$99 \div 4 = 24\textrm{ R }3$

The remainder is 3, so

$i^{99} = i^{3}$

Consequently, using the Product of Powers Rule:

$i^{99} = i^{2} \cdot i = -1 \cdot i = -i$

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Example Question #26 : Number Theory

Let be a complex number. $\overline{x}$ denotes the complex conjugate of .

$x+\overline{ x} = 20$ and $x \overline{ x} = 200$ .

How many of the following expressions could be equal to ?

(a) 10 + 10i

(b) 10 - 10i

(d) -10 - 10i

Possible Answers:

One

None

Four

Two

Three

Correct answer:

Two

Explanation:

is a complex number, so x = a+ bi for some real a ,b ; also, $\overline{x} = a-bi$ .

Therefore,

$x+\overline{ x} = (a+ bi) + (a-bi) = a+ a + bi - bi = 2a$

Substituting:

$x+\overline{ x} = 20$

2a = 20

a = 10

Therefore, we can eliminate choices (c) and (d).

Also, the product

$x \overline{ x} = (a+bi)(a-bi) = a^{2} +b^{2}$

Setting $x \overline{ x} = 200$ and substituting 10 for , we get

$10^{2} +b^{2} = 200$

$100 +b^{2} = 200$

$b^{2} = 100$

$b = \pm 10$

Therefore, either x = 10 + 10i or x = 10 - 10i - making two the correct response.

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SAT II Math II : Real and Complex Numbers

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #11 : Sat Subject Test In Math Ii

Example Question #11 : Real And Complex Numbers

Example Question #21 : Number Theory

Example Question #11 : Real And Complex Numbers

Example Question #11 : Real And Complex Numbers

Example Question #22 : Number Theory

Example Question #23 : Number Theory

Example Question #24 : Number Theory

Example Question #26 : Number Theory

All SAT II Math II Resources

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